Shooting from opposite side of the field

[Brandon_L]

After hearing the balls were foam, I was worried about how they would handle being launched with air resistance (as a foamy ball would barley go anywhere, like throwing a un-crumpled piece of paper). After I got my hands on one though, it seems a lot better then I thought it would be.

Anyway, my question would be do you think a shooter powerful enough to score from the other side of the field (or up against the bump on the other side, at least) be possible? and is scoring from the other side allowed? (I saw nothing in the rules)

[davidthefat]

Yes, I believe it is possible. However, to actually score, I believe it is a one in a thousand chance.

[pfreivald]

I think you'll see very competitive teams do it, yes.

[Djur]

I ran some quick calculations, and in order to score from an opposite corner, the ball would need to have an exit velocity of about 42 feet/second at a 45* angle.

[Brandon_L]

Originally Posted by Djur

I ran some quick calculations, and in order to score from an opposite corner, the ball would need to have an exit velocity of about 42 feet/second at a 45* angle.

corner, as in where human players put balls into play? what about somewhere closer but still on the other side?

What formulas did you use?

And thanks!

[Djur]

I used this trajectory calculator. It's pretty neat. This one may be useful, too.

Yeah, I meant the corner where the ball return is, but I can't see the velocity from the center of the opposite side being much different, maybe 40 ft/sec.

[shuhao]

don't know how accurate your estimation will be. at 42 feet per second (12m/s) initial you will run into about 1.2 Newton of drag force.. which translates to approximately 3.5m/s^2 of initial deceleration.

[Justin Montois]

Originally Posted by pfreivald

I think you'll see very competitive teams do it, yes.

By "do it" do you mean just shooting from the far side of the field or actually scoring...

[CalTran]

I've been doing some math, and 6inches from the edge and 6 inches from the very far back wall of the lane, it's a 55.0568foot shot to the goal, not including the distance for an angle to the top of the goal. The extra distance it has to go, assuming the cannon/turret is at the very top of the 60" limit, is negligible (About an inch or so). At any rate, a 55 foot shot is going to be quite difficult.From the research I've just been doing, it's 42' to half court, then you'd add another 13' to that.

[krazyman1013]

Look at the Peguineers robots from 2006.

[NorviewsVeteran]

On a less scientific note, I do remember nearly being pegged in the face in 2006 by an over propelled poof ball (I kept wondering why it was getting bigger and bigger...) while standing past the driver wall. While not quite the same as the basketballs this year, they can carry for some distance.

[shuhao]

I don't think that's possible this year.. in order for this thing to go far, it needs to be at 45 degrees and have a really fast speed.. with this weight... any speed above 15m/s will slow down very quickly..

will mock up a simulation in the morning in python with an iterative estimator

[Djur]

Ran through Wikipedia and Google quickly and came up with this:
Drag force = (1/2) * p * v^2 * CoD * A

• p = air density
  v = velocity
  CoD = Coefficient of Drag
  A = relative area

Fill in with measurements:
Drag force = (1/2) * 1.204 kg/m^3 * (12.8016 m/s)^2 * 0.4 * 0.0162146393 m^2 = 2.010 Newtons (approx. 0.452 pounds).

Fairly negligible if my math is right. Mathematically, you could figure out how much force is needed to have the ball travel at a speed without regarding drag, then just tack the drag force onto that. Making something to force a ball to go at that speed will be hard, though.

EDIT1: Fixed a stupid math error (00:49)
EDIT2: Check out this post on drag etc. (01:13)

[Gray Adams]

Originally Posted by Djur

Ran through Wikipedia and Google quickly and came up with this:
Drag force = (1/2) * p * v^2 * CoD * A

• p = air density
  v = velocity
  CoD = Coefficient of Drag
  A = relative area

Fill in with measurements:
Drag force = (1/2) * 1.204 kg/m^3 * (12.8016 m/s)^2 * 0.4 * 0.638 m^2 = 25.192 Newtons (approx. 5.337 pounds).

So yeah, that's gonna be a bit tough. Mathematically, you could figure out how much force is needed to have the ball travel at a speed without regarding drag, then just tack the drag force onto that. Making something to force a ball to go at that speed will be hard, though.

I haven't really done any math, but putting a wheel directly on a CIM spins terrifyingly fast, and has a sound to match.

[davidthefat]

Originally Posted by Gray Adams

I haven't really done any math, but putting a wheel directly on a CIM spins terrifyingly fast, and has a sound to match.

You can even gear it down and have it go faster.